So, an eigenstate of an observable always has all corresponding
\(V_{\!\scriptscriptstyle \hat{\beta }_n}\)  being zero, and degenerate eigenstates for
an observable have identical noncommutative values. Moreover,
\(\hat{\beta }=r\hat{I}\)  gives \(f=r\) , have the noncommutative value
behaving essentially as a commutative classical real number
value. Note that the matrix element \((\hat{\beta })^m_n\)  can be
expressed in terms of \(f_{\!\scriptscriptstyle \hat{\beta }}\) , \(V_{\!\scriptscriptstyle \hat{\beta }_n}\)  and
\({\tilde{k}}_{\!\scriptscriptstyle \hat{\beta }_{\bar{m}n}} \equiv \partial _{{n}} \partial _{\bar{m}} f_{\!\scriptscriptstyle \hat{\beta }}\)  [1]},
hence the full sequence for the noncommutative value can be
obtained from a given expectation value function \(f_{\!\scriptscriptstyle \hat{\beta }}\) 
on the projective Hilbert space without knowing a priori
the explicit operator form of the \(\hat{\beta }\) . In fact, one can
check if a function \(f(z_n, \bar{z}_n)\)  is indeed an \(f_{\!\scriptscriptstyle \hat{\beta }}\) 
without knowing \(\hat{\beta }\)  [2]}, [3]}. Moreover, the
classical value \(r\)  as a constant noncommutative value has also
\({\tilde{k}}_{\!\scriptscriptstyle \hat{\beta }_{\bar{m}n}}=0\) .
